Latest Elitmus Aptitude Question SOLUTION: How many no can be formed using digits 1,2,3,4,5,6,7,8,9 such that they are in increasing order(eg. 12345 ,578)

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(#M40020014) ELITMUS question Keep an EYE

How many no can be formed using digits 1,2,3,4,5,6,7,8,9 such that they are in increasing order(eg. 12345 ,578)

Asked In Elitmus Pratyush (10 years ago)

Unsolved Read Solution (9)

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vaibhav sharma  (10 years ago)

If you start with 9, you can only do the one digit. If you start with 8, you can do 8, or 89. If you start with 7, you can do 7, 78, 79, or 789. If you start with 6, you can do 6, 67, 68, 69, 678, 679, 689, or 6789. I'll do one more. If you start with 5, you can do 5, 56, 57, 58, 59, 567, 568, 569, 578, 579, 589, 5678, 5679, 5689, 5789, 56789.
so when we start with 9,8,7,6,5...upto 1 then we have 1,2,4,8,16,32,....256 ways
respectively so add all these ways we get 2^0+2^1+...2^8=511

Is this solution Helpfull? Yes (24) | No (4)  

nitin gupta  (10 years ago)

2digit+3digit+.....+9dgit

9*8/2+ 9*8*7/6+ 9*8*7*6/24+ 9*8*7*6*5/120+.....=502

Is this solution Helpfull? Yes (7) | No (2)  

VIKAS SINGH  (10 years ago)

9c2+9c3+9c4+9c5+9c6+9c7+9c8+9c9=502

Is this solution Helpfull? Yes (4) | No (3)  

SACHIN AGGARWAL  (10 years ago)

9c0+9c1+9c2+9c3+9c4+9c5+9c6+pc7+9c8+9c9= 2^9-1 =511.

Is this solution Helpfull? Yes (4) | No (1)  

Sachin Kataria  (10 years ago)

9*8/2 + 8*7*9/2 + 7*6*9/2 + 6*5*9/2 + 5*4*9/2 + 4*3*9/2 + 3*2*9/2 + 2*1*9/2 = 766

Is this solution Helpfull? Yes (2) | No (8)  

Mohit Bajaj  (10 years ago)

8*9/2+7*8*9/2+6*7*8*9/2+5*6*7*8*9/2+4*5*6*7*8*9/2+3*4*5*6*7*8*9/2+
2digi+3digits+4digits +5 digits +6 digits +7 digits +
+2*3*4*5*6*7*8*9/2+1*2*3*4*5*6*7*8*9/2
+8 digits +9 digits

Is this solution Helpfull? Yes (1) | No (3)  

siddanth  (10 years ago)

answer 520

Is this solution Helpfull? Yes | No (9)  

ranjeet singh chauhan  (10 years ago)

Could you elaborate your answer.

Is this solution Helpfull? Yes | No (1)  

Manish  (7 years ago)

use simple formula for every digit as-2^9-d where d is digit
2^9-1+2^9-2+2^9-3............2^9-9=256+128+64+32+16+8+4+2+1=511

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